Results 1 to 10 of 10

Math Help - Rhombus

  1. #1
    Member
    Joined
    May 2007
    Posts
    150

    Rhombus

    The diagonals of a rhombus ABCD meet at M. Angle DAB is 60 degress. P is the midpoint of AD. G is the intersection of PC and MD. AP = 8.

    Find MD MC MG CG and GP


    so far i got
    Md - 13.86
    MC - 8
    MG 4.62
    CG - 9.24

    im not sure if im right tho
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,904
    Thanks
    329
    Awards
    1
    Quote Originally Posted by stones44 View Post
    The diagonals of a rhombus ABCD meet at M. Angle DAB is 60 degress. P is the midpoint of AD. G is the intersection of PC and MD. AP = 8.

    Find MD MC MG CG and GP


    so far i got
    Md - 13.86
    MC - 8
    MG 4.62
    CG - 9.24

    im not sure if im right tho
    The line BD cuts the rhombus in half, bisecting the angles CDA and ABC. Since these angles are (1/2)*(360 - 2*60) = 120 degrees, triangle ABD is an equilateral triangle. Thus MD = 16.

    Triangle ACD is an isosceles triangle, with the two equal sides (CD and DA) of 16, and vertex angle 120 degrees, so AC will be
    AC = sqrt{CD^2 + AD^2 - 2*CD*AD*cos(12)} <-- By the Law of Cosines

    AC = 27.7128

    Try working it from here.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2007
    Posts
    150
    ooooh

    thanks a bunch....i never see those obvious equilateral or isoscles triangles

    and i think you mean MD is 8 because it is half of BD (which is 16)


    but: that only helps me get MD and MC...I still have a hard time finding the other three
    Last edited by stones44; May 7th 2007 at 03:12 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,904
    Thanks
    329
    Awards
    1
    Quote Originally Posted by stones44 View Post
    and i think you mean MD is 8 because it is half of BD (which is 16)
    Yes.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,904
    Thanks
    329
    Awards
    1
    I'm not going to do the calculations, but it's not that hard if you go step by step.

    Do CG first.

    We know that CD = 16 and DP = 8 and angle CDP = 120 degrees. Thus
    CP^2 = CD^2 + DP^2 - 2*CD*DP*cos(120)
    by Law of Cosines. (Triangle CDP)

    Now we can find angle PCD:
    sin(angle PCD)/8 = sin(angle CDP)/CP
    by Law of Sines. (Triangle CDP)

    Now we can find angle CGD:
    angle CGD = 180 - angle CDG - (1/2)*120<-- Angle CDG is 1/2 of 120 since BD is the diagonal of the rhombus.
    since the sum of the interior angles of a triangle is 180 degrees. (Triangle CGD)

    Now we can find CG:
    sin(angle CGD)/16 = sin(60)/CG
    by Law of Sines (Triangle CGD)

    Now do MC:

    sin(60)/MC = sin(30)/8 <-- MD = 8
    by Law of Sines (Triangle CMD)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2006
    Posts
    74
    to find the other length you shoul notice that CP amd DM are medians in triangle ADC, so G is the center of mass of that triangle.
    So, we have GM=1/3 MD,
    CG=2/3PC
    PG=1/3CP
    Use Cosine law in triangle CPA to find CP.
    Using sine law in this triangle, after you find the sides will give you the angle ACP.
    Hope this helps,
    Alina
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2007
    Posts
    150
    thanks..this all helped

    is there a way to do it without cosine or sine? i believe there is.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,904
    Thanks
    329
    Awards
    1
    Quote Originally Posted by stones44 View Post
    thanks..this all helped

    is there a way to do it without cosine or sine? i believe there is.
    I doubt it. But if you figure one out I'd be interested in seeing it.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    May 2007
    Posts
    150
    i know there is.....the questions is in a subject before trig is taught...hmmm if i can find PC without trig, i can get everything else
    Last edited by stones44; May 8th 2007 at 02:23 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    May 2007
    Posts
    150
    YESSS I DID IT!!!

    ok, so using the median theroum i can find that since DM is 8, and MG is 1/3 of DM (because it is a median of triangle ADC) i find mg is 2.666666.......so then i use pythagoras and do MG^2*MC^2 (MC, which i found previously) and this gets me CG....then i know that cg is 2/3 of PC becuase PC is also a median of triangle ADC....so, PG is 1/3 of PC....

    NICE!!!!!!!!!!!!!!!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. rhombus
    Posted in the Geometry Forum
    Replies: 1
    Last Post: February 11th 2010, 02:55 AM
  2. rhombus
    Posted in the Geometry Forum
    Replies: 3
    Last Post: January 7th 2010, 07:18 AM
  3. help rhombus
    Posted in the Geometry Forum
    Replies: 1
    Last Post: October 12th 2007, 01:04 PM
  4. Rhombus
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 29th 2007, 03:58 PM
  5. Rhombus
    Posted in the Geometry Forum
    Replies: 3
    Last Post: May 5th 2007, 08:31 AM

Search Tags


/mathhelpforum @mathhelpforum